0:00
Hi there.
00:01
So for this problem we have the situation that is shown in this figure so water is flowing from an open tank and we know that the elevation of the point one we're gonna call that the height 1 that is 10 meters and we're also given that the elevation of points 2 and 3 are the same and they are 2 meters.
00:34
We're also given the information about the cross -ceptional area of the points 2 and 3.
00:41
So the creusceptional area of 2 of 0 .2 is 0 .48 meters squared and the cross -ceptional area of the 0 .3 is 0 .0 .0.
01:04
16 meters square.
01:08
Now, the problem also told us that the area of the tank is very large compared to the other's perceptional area of the pipe.
01:19
So what we need to do is to use bernoulli's equation.
01:25
To the first thing that we need to compute is the discharge rate in cubic meters per second.
01:33
So what we need to calculate.
01:35
Is the discharge change that is called q per meters per cubic meters per second.
01:53
So that is.
01:56
So in this case, that this church rate q is defined as the correctional area 3 times the velocity at the point 3.
02:14
Because that's and that is the area from where the water is coming out of this tank the area in here.
02:24
So what we need to calculate first is the speed three of the of the water at that point.
02:35
So what we are going to apply is the bernoulli's equation.
02:42
So we know that the pressure at, point one plus and we also have a term related to this speed so that will be one half of the one half of the density of water in this case times the velocity at point one squared these times a term related with the potential with the potential energy that is the density of water times the acceleration due to gravity times the height want of the point one.
03:26
And this should be equal to the pressure at 0 .3 plus a term related to the speed at that point.
03:38
So that this is 3 squared plus the pressure times acceleration.
03:44
So we obtain this.
03:46
Now we know that seems at the point 3 and at the point 1, both are in contact with the atmosphere.
03:56
So p1 and p3 corresponds to the atmospheric pressure, so we can console those two.
04:04
And also as the creceptional area of the tank, of the tank where the point one is located at, is very large in comparisons to the other perceptional area.
04:19
We can assume that the velocity, of this point is just very tiny in comparison to the other velocities.
04:31
So we can assume that this is zero.
04:34
So we are left with these two terms.
04:39
So we can also eliminate the dependence on the density of water because they are in all of these terms.
04:51
So we will have, we can pass this to the order.
04:55
Side so we will have g times h1 minus h3 the height the heights those two heights for points one in two and three and this times two and taking the square root of this will give us the velocity at point three or the speed of point three now we substitute all of these values because we know all of then we know that the acceleration due to gravity is 9 .8 meters per second square.
05:29
We know that the height, the elevation of the point one, is 10 meters, and the elevation of the point three is 2 meters.
05:39
So we need to subtract those two values.
05:44
So plurting this into the calculator, we obtain that the speed of the water at 0 .3 has a value of 12...